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How many threedigit numbers ABC, in which A, B, and C are each digits [#permalink]
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08 Jul 2016, 21:44
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How many threedigit numbers ABC, in which A, B, and C are each digits, satisfy the equation 2B = A + C? A. 33 B. 36 C. 41 D. 45 E. 50
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Re: How many threedigit numbers ABC, in which A, B, and C are each digits [#permalink]
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09 Jul 2016, 01:25
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first 3 digit number which qualifies for 2B=A+C > 111 next is 123 next is 135 gap of 12 900 3 digit numbers so 900/12=50
E?



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Re: How many threedigit numbers ABC, in which A, B, and C are each digits [#permalink]
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09 Jul 2016, 03:12
Possible solutions are Diff of 0  111,222,........,999 = 9 Diff of 1  123, 234, ....., 789 =7x2 = 14 Diff of 2  135, 246.......579 = 5x2 = 10 Diff of 3  147,258,369 = 3x2 = 6 Diff of 4  159 = 1x2 = 2 Last 210, 420,630,840 = 4
So total is 9+14+10+6+2+4 = 45.
Not a good way to solve. Experts please
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How many threedigit numbers ABC, in which A, B, and C are each digits [#permalink]
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09 Jul 2016, 05:34
2B = A+C Middle number can only be 2, 4,6,8,10,12,14,16,18 now when middle number is 2 , you can have only 1 combination A =1 and C =1
when middle is 4 you can have two pairs 1 ,3 and 2,2 when middle is 6 you can have three so basically 1+2+3+4+5+6+7+8+9 = 45 D is the answer
I ma confused about one thing , why we are not required to consider op pair when we have 4 at middle we can have (1,3 ) (2,2)and (3,1) when we have 6 at middle we can have we have 1,5 2,4 3,3 4,2 51 another series is 1+3+5+...17 == 81 , which is not answer



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Re: How many threedigit numbers ABC, in which A, B, and C are each digits [#permalink]
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10 Jul 2016, 02:21
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2B = A + C Let A=C therefore 2A = 2B therefore A=B=C is part of solution. There are 9 such occurrences
now let A = 1, possible values of c =3,5,7,9 A = 2, possible values of c =4,6,7,0 A = 3, possible values of c =1,5,7,9 A = 4, possible values of c =2,6,8,0 ... so on... There are 9 such rows = 9 * 4 = 36 Total = 36 + 9 = 45 D



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How many threedigit numbers ABC, in which A, B, and C are each digits [#permalink]
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07 Oct 2016, 22:20
I just did manual combinations.. Though I got it right, it took lot of time.. ABC Start with number 9 for B..
A9C > we have to find combinations of A and C such that A+C=18 > this yields only one combination > 999 A8C > we have to find combinations of A and C such that A+C=16 > this yields three combinations > 888, 987, 799 .. .. .. ..
Do the same for all digits and make the list B num of combinations 9 1 8 3 7 5 6 7 5 9 4 8 3 6 2 4 1 2 0 0  Total  45 (add all above numbers in red color)
Answer choice D is correct.. But this procedure takes lot of time...
Is there any quick solution?



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How many threedigit numbers ABC, in which A, B, and C are each digits [#permalink]
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28 Oct 2016, 02:33
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2B=A+C is the same as B=(A+C)/2. That is B is the arithmetic mean of the other 2 digits. This can be achieved only in case, when parity of both A and C is the same. So we have odd + odd or even + even. Let's look at first case. We have 5 ways to choose odd number for A and same 5 ways to choose odd number for C  total 5*5=25. In the case of even numbers, we can't choose 0 for A (otherwise it won't be 3 digit number) so we have 4*5=20. Those cases are independent so in total we have 45 ways. Answer is D.



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How many threedigit numbers ABC, in which A, B, and C are each digits [#permalink]
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28 Oct 2016, 15:53
alpham wrote: How many threedigit numbers ABC, in which A, B, and C are each digits, satisfy the equation 2B = A + C?
A. 33 B. 36 C. 41 D. 45 E. 50 If we assume numbers that work are distributed equally through the 9 100s blocks, only answers B and D are multiples of 9. If we choose any 100s block, say the 500s, and start at 555, which we know will work, and count forward and backward by intervals of 12 until numbers no longer satisfy, we will have 5 numbers that work: 555,567,579,543,531. 5*9=45 D



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Re: How many threedigit numbers ABC, in which A, B, and C are each digits [#permalink]
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13 Nov 2017, 05:42
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vitaliyGMAT wrote: 2B=A+C is the same as B=(A+C)/2. That is B is the arithmetic mean of the other 2 digits. This can be achieved only in case, when parity of both A and C is the same. So we have odd + odd or even + even. Let's look at first case. We have 5 ways to choose odd number for A and same 5 ways to choose odd number for C  total 5*5=25. In the case of even numbers, we can't choose 0 for A (otherwise it won't be 3 digit number) so we have 4*5=20. Those cases are independent so in total we have 45 ways. Answer is D. Alternatively, we are given that 2B = A + C So whatever A and C are, they add up to give an even sum. Hence either A and C are both even or both odd. Now proceed as given above.
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How many threedigit numbers ABC, in which A, B, and C are each digits [#permalink]
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13 Nov 2017, 15:23
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alpham wrote: How many threedigit numbers ABC, in which A, B, and C are each digits, satisfy the equation 2B = A + C?
A. 33 B. 36 C. 41 D. 45 E. 50 1.Interpret what is given so that we get basic info. 2B = A + C implies that A and C are both odd or both even 2. When A and C are both odd, A can be 1,3,5,7,9 and for each of these B can be 1,3,5,7,9 for a total of 25 cases 3. When A and C are both even A can be 2,4,6,8 and for each of these B can be 0,2,4,6,8 for a total of 20 cases 4. So the answer is 25+20= 45 cases
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Re: How many threedigit numbers ABC, in which A, B, and C are each digits [#permalink]
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16 Nov 2017, 13:32
mbaprep2016 wrote: 2B = A+C Middle number can only be 2, 4,6,8,10,12,14,16,18 now when middle number is 2 , you can have only 1 combination A =1 and C =1
when middle is 4 you can have two pairs 1 ,3 and 2,2 when middle is 6 you can have three so basically 1+2+3+4+5+6+7+8+9 = 45 D is the answer
I ma confused about one thing , why we are not required to consider op pair when we have 4 at middle we can have (1,3 ) (2,2)and (3,1) when we have 6 at middle we can have we have 1,5 2,4 3,3 4,2 51 another series is 1+3+5+...17 == 81 , which is not answer I also have the same doubt. Can some expert help here?
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How many threedigit numbers ABC, in which A, B, and C are each digits [#permalink]
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16 Nov 2017, 16:33
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GMATisLovE wrote: mbaprep2016 wrote: 2B = A+C Middle number can only be 2, 4,6,8,10,12,14,16,18 now when middle number is 2 , you can have only 1 combination A =1 and C =1
when middle is 4 you can have two pairs 1 ,3 and 2,2 when middle is 6 you can have three so basically 1+2+3+4+5+6+7+8+9 = 45 D is the answer
I ma confused about one thing , why we are not required to consider op pair when we have 4 at middle we can have (1,3 ) (2,2)and (3,1) when we have 6 at middle we can have we have 1,5 2,4 3,3 4,2 51 another series is 1+3+5+...17 == 81 , which is not answer I also have the same doubt. Can some expert help here? Hi, We have 2B=A+C as follows: when middle number is 1 we have the pairs (1,1) (2,0) When middle number is 2 we have the pairs (2,2 ) (3,1) (1,3 ) (4,0) When middle number is 3 we have the pairs (3,3) (4,2), (2,4) (5,1) (1,5) (6,0) When middle number is 4 we have the pairs (4,4 ) (5,3) (3,5) (6,2) (2,6) (1,7) (7,1) (8,0) When middle number is 5 we have the pairs (5,5) (6,4,) (4,6) (7,3) (3,7) (8,2) (2,8) (9,1) (1,9) When middle number is 6 we have the pairs (6,6 ) (7,5) (5,7) (4,8 ) (8,4 ) (9,3) (3,9 ) When middle number is 7 we have the pairs (7,7) (8,6) (6,8) (9,5) (5,9) When middle number is 8 we have the pairs (8,8 ) (9,7) (7,9) When middle number is 9 we have the pairs ( 9,9) for a total of 45 cases
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How many threedigit numbers ABC, in which A, B, and C are each digits [#permalink]
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29 Dec 2017, 07:41
Given : ABC is three digit number, so A > 0 A + C = 2B => A + C = even
case 1: both A and C are odd odd digits: {1,3,5,7,9} = there are 5 * 5 possible permutations => total = 25
case 2: both A and C are even even digits : {0,2,4,6,8} => possible digits for A : {2,4,6,8} = 4 possible digiits for C: {0,2,4,6,8} = 5 total permutations : 4 * 5 = 20
so number of such three digit number = 25 + 20 = 45 => (D)




How many threedigit numbers ABC, in which A, B, and C are each digits
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